= Solution
Measure $\theta$ from the upward vertical, let $\mathbf e_3$ point from the support along the <symmetry> axis, and use the support as the reference point for all <moments of inertia>. Choose $\mathbf e_2=(\widehat{\mathbf z}-\cos\theta\mathbf e_3)/\sin\theta$ and $\mathbf e_1=\mathbf e_2\times\mathbf e_3$. These form an instantaneous principal-axis frame; its transverse axes need not spin with marks painted on the body. The three infinitesimal <Euler angles> rotations add as vectors:
$$
\boldsymbol\omega=\dot\theta\mathbf e_1+\dot\phi\widehat{\mathbf z}+\dot\chi\mathbf e_3.
$$
Resolving the vertical direction gives
$$
\boxed{\boldsymbol\omega=(\dot\theta,\dot\phi\sin\theta,\dot\phi\cos\theta+\dot\chi),\qquad
\mathbf J=(I_1\dot\theta,I_1\dot\phi\sin\theta,I_3[\dot\phi\cos\theta+\dot\chi]).}
$$
A convention matters here: in genuinely material-fixed transverse axes, the first two components are rotated by the spin angle $\chi$. For one conventional choice they are $\dot\theta\cos\chi+\dot\phi\sin\theta\sin\chi$ and $-\dot\theta\sin\chi+\dot\phi\sin\theta\cos\chi$. Since $I_1=I_2$, that rotation changes neither <kinetic energy> nor the axial and vertical conserved momenta. The simplified displayed components use the instantaneous transverse frame just defined.<Classical Dynamics by David Tong/The Motion of Rigid Bodies>[§3.5.2, angular-velocity decomposition and body-frame components; §3.6, symmetric-top conserved momenta]{ref}
For a frictionless fixed support, the support does no work and gravity is conservative, so the total <energy> is constant. The <heavy symmetric top Lagrangian> is
$$
\mathcal L=\frac{I_1}{2}(\dot\theta^2+\dot\phi^2\sin^2\theta)
+\frac{I_3}{2}(\dot\chi+\dot\phi\cos\theta)^2-mgh\cos\theta.
$$
The two <cyclic coordinates> $\chi,\phi$ give
$$
p_\chi=J_3=I_3(\dot\chi+\dot\phi\cos\theta),\qquad
p_\phi=J_z=I_1\dot\phi\sin^2\theta+J_3\cos\theta.
$$
Equivalently, gravitational <torque> has no component along the vertical or <symmetry> axis. Eliminate $\dot\phi=(J_z-J_3\cos\theta)/(I_1\sin^2\theta)$ and the axial spin from $E=T+V$:
$$
\boxed{E=\frac{I_1}{2}\dot\theta^2+U_{\rm eff}(\theta),\qquad
U_{\rm eff}=\frac{(J_z-J_3\cos\theta)^2}{2I_1\sin^2\theta}
+mgh\cos\theta+\frac{J_3^2}{2I_3}.}
$$
This is the <heavy symmetric top reduction> at fixed $J_3,J_z$. A constant inclination must satisfy the reduced <equation of motion> $I_1\ddot\theta=-U_{\rm eff}'$, not merely have zero instantaneous <velocity>. Differentiation at fixed conserved momenta gives, for $0<\theta<\pi$ and $\Omega=\dot\phi$,
$$
U_{\rm eff}'=\sin\theta\,[J_3\Omega-I_1\Omega^2\cos\theta-mgh].
$$
Therefore <uniform precession of a heavy symmetric top> requires
$$
\boxed{I_1\cos\theta\,\Omega^2-J_3\Omega+mgh=0.}
$$
For a prescribed upward inclination $0<\theta<\pi/2$, real <precession> rates exist only if the quadratic <discriminant> is nonnegative:
$$
\Omega_\pm=\frac{J_3\pm\sqrt{J_3^2-4I_1mgh\cos\theta}}{2I_1\cos\theta},\qquad
\boxed{|J_3|_{\min}(\theta)=2\sqrt{I_1mgh\cos\theta}.}
$$
For positive spin the absolute-value sign can be dropped. The two roots are the slow and fast branches and coalesce at the threshold. The familiar upright-limit critical magnitude is $2\sqrt{I_1mgh}$; strict inequality gives <gyroscopic stabilization of an inverted symmetric top>. This limiting stability condition should not be confused with a positive minimum over every possible inclination. As $\theta$ approaches $\pi/2$ from below, the fixed-inclination threshold approaches zero. At the horizontal angle $\Omega=mgh/J_3$ for $J_3\ne0$; below the horizontal the quadratic admits real solutions even at $J_3=0$. The stated minimum therefore needs the inclination convention made explicit.
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